Deriving charge for Noether current in free complex scalar field QFT


by Azelketh
Tags: charge, complex, current, deriving, field, free, noether, scalar
Azelketh
Azelketh is offline
#1
Nov14-12, 11:00 AM
P: 36
1. The problem statement, all variables and given/known data
Hi I a attempting to derive the expression for the conserved Noether charge for a free complex scalar field.
The question I have to complete is: " show, by using the mode expansions for the free complex scalar field, that the conserved Noether charge (corresponding to complex phase rotations) is given by
[tex] Q= \int \frac{\partial^3 p}{(2\pi)^3} ( a_p^* a_p - b_p^* b_p)
[/tex]
Where [tex]a_p, a_p^*[/tex] and [tex] b_p, b_p^*[/tex] are the creation and annihilation operators for 2 kinds of particles respectively"
2. Relevant equations

The Lagrangian for a free complex scalar field is;
[tex] L= \frac{1}{2} \partial_μ \psi \partial^μ \psi- \frac{1}{2}m^2\psi^2 [/tex]

I have recognized that there is a symmetry of[tex]\psi → e^{ i\alpha } \psi [/tex]

which leads to a Noether current of
[tex] j_μ = i \partial _μ \psi^* \psi - i \partial _μ \psi \psi^* [/tex]
[tex] Q= \int \partial ^3 x j_0 = i \int \partial ^3 x ( \partial _0 \psi^* \psi - \partial _0 \psi \psi^* ) [/tex]
[tex] Q=i \int \partial ^3 x ( \psi \dot{\psi^*} - \psi^* \dot{\psi} )[/tex]
where
[tex] \dot{\psi} = \pi ^*[/tex]
[tex] \dot{\psi^*} = \pi[/tex]
so [tex] Q=i \int \partial ^3 x ( \psi \pi - \psi^* \pi ^* )[/tex]
And the mode expansions;
[tex] \psi = \int \frac{\partial^3 P}{(2\pi)^3} \frac{1}{\sqrt{2 ω_p}}( a_p e^{ip.x } + b_p^* e^{-ip.x}) [/tex]
[tex] \psi^* = \int \frac{\partial^3 P}{(2\pi)^3} \frac{1}{\sqrt{2 ω_p}}( a_p^* e^{-ip.x} + b_p e^{ip.x }) [/tex]
[tex] \pi = i\int \frac{\partial^3 P}{(2\pi)^3} \sqrt{ \frac{ω_p}{2}}( a_p^* e^{-ip.x } - b_p e^{ip.x}) [/tex]
[tex] \pi^* = -i\int \frac{\partial^3 P}{(2\pi)^3} \sqrt{\frac{ω_p}{2}}( a_p e^{ip.x } - b_p^* e^{-ip.x}) [/tex]

3. The attempt at a solution

I have then plugged these into the Q equation;
[tex] Q=i \int \partial ^3 x [ \int \frac{\partial^3 P}{(2\pi)^3} \frac{1}{\sqrt{2 ω_p}}( a_p e^{ip.x } + b_p^* e^{-ip.x})
i\int \frac{\partial^3 P}{(2\pi)^3} \sqrt{ \frac{ω_p}{2}}( a_p^* e^{-ip.x } - b_p e^{ip.x})
- \int \frac{\partial^3 P}{(2\pi)^3} \frac{1}{\sqrt{2 ω_p}}( a_p^* e^{-ip.x} + b_p e^{ip.x }) (-i)\int \frac{\partial^3 P}{(2\pi)^3} \sqrt{\frac{ω_p}{2}}( a_p e^{ip.x } - b_p^* e^{-ip.x}) ][/tex]
so factoring out the i's and 2 pi's from the integrals;
[tex] Q=\frac{i^2}{(2\pi)^6} \int \partial ^3 x [ \int \partial^3 P \frac{1}{\sqrt{2 ω_p}}( a_p e^{ip.x } + b_p^* e^{-ip.x})
\int \partial^3 P \sqrt{ \frac{ω_p}{2}}( a_p^* e^{-ip.x } - b_p e^{ip.x})
+ \int \partial^3 P \frac{1}{\sqrt{2 ω_p}}( a_p^* e^{-ip.x} + b_p e^{ip.x }) \int \partial^3 P \sqrt{\frac{ω_p}{2}}( a_p e^{ip.x } - b_p^* e^{-ip.x}) ][/tex]
It's at this point I'm utterly stuck on how to procede, I did think that could cancel the sqrt ω terms but they depend on p: [tex] ω_p = \sqrt{P^2 + m^2} [/tex]
so as they are inside integrals dependent on P they can't be cancelled.

I am at an utter loss how to proceed from here though. If anyone can offer any pointers or assistance is would be greatly appreciated.
EDIT: made a sign error in exponential's for [tex] \psi [/tex] and [tex] \psi^* [/tex]
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TSny
TSny is offline
#2
Nov14-12, 12:15 PM
HW Helper
Thanks ∞
PF Gold
P: 4,513
You can try changing the order of integration and doing the x integration before the p integrations.
Azelketh
Azelketh is offline
#3
Nov14-12, 01:27 PM
P: 36
Ok, I see how you can change the order of the integration, and changing one of the P's to r to make it clearer to me, to get :
[tex]
Q=\frac{-1}{(2\pi)^6} \int \partial ^3 x [ \int \partial^3 P \frac{1}{\sqrt{2 ω_p}}( a_p e^{ip.x } + b_p^* e^{-ip.x})
\int \partial^3 r \sqrt{ \frac{ω_r}{2}}( a_r^* e^{-ir.x } - b_r e^{ir.x})
+ \int \partial^3 P \frac{1}{\sqrt{2 ω_p}}( a_p^* e^{-ip.x} + b_p e^{ip.x }) \int \partial^3 r \sqrt{\frac{ω_r}{2}}( a_r e^{ir.x } - b_r^* e^{-ir.x}) ]
[/tex]
[tex]
Q=\frac{-1}{(2\pi)^6} \int \partial ^3 x [ \int \partial^3 P \sqrt{\frac{ω_r}{ ω_p}}
( a_p e^{ip.x } + b_p^* e^{-ip.x})
\int \partial^3 r ( a_r^* e^{-ir.x } - b_r e^{ir.x})
+ \int \partial^3 P \sqrt{\frac{ω_r}{ ω_p}}( a_p^* e^{-ip.x} + b_p e^{ip.x }) \int \partial^3 r ( a_r e^{ir.x } - b_r^* e^{-ir.x}) ]
[/tex]
[tex]
Q=\frac{-1}{(2\pi)^6} ( \int \partial^3 P \int \partial^3 r \int \partial ^3 x \sqrt{\frac{ω_r}{ ω_p}}( a_p e^{ip.x } + b_p^* e^{-ip.x} )
( a_r^* e^{-ir.x } - b_r e^{ir.x})
+ \int \partial^3 P \int \partial^3 r \int \partial ^3 x \sqrt{\frac{ω_r}{ ω_p}}( a_p^* e^{-ip.x} + b_p e^{ip.x })( a_r e^{ir.x } - b_r^* e^{-ir.x}) )
[/tex]

[tex]
Q=\frac{-1}{(2\pi)^6} ( \int \partial^3 P \int \partial^3 r \int \partial ^3 x \sqrt{\frac{ω_r}{ ω_p}}
( a_p a_r^* e^{i( p - r ).x} -a_p b_r e^{i( p + r).x} + b_p^* a_r^* e^{-i( p + r ).x} -b_p^* b_r e^{i( r - p ).x} )
+ \int \partial^3 P \int \partial^3 r \int \partial ^3 x \sqrt{\frac{ω_r}{ ω_p}}
( a_p^* a_r e^{i( r - p ).x} -a_p^* b_r^* e^{-i( P + r ).x} + b_p a_re^{i( p + r).x} - b_p b_r^* e^{i( p - r ).x} )
[/tex]

I'm not sure how to go about integrating the x components;
I've never integrated anything over [tex] \partial^3 x [/tex] and am struggling to find anything that can help explain it.
would;
[tex]
\int \partial ^3 x a_p^* a_r e^{ i(p +r).x} = \frac{1}{ (i(p+r))^3}a_p^* a_r e^{ i(p +r).x}
[/tex]
or as its a dot product in the exponential,
[tex]
\int \partial ^3 x a_p^* a_r e^{ i(p +r).x} = \int \partial ^3 x a_p^* a_r e^{ i( (p1 +r1)x1 + (p2+r2)x2 +(p3+r3)x3 )}
[/tex]
[tex]
= \frac{1}{i^3(p1+r1)(p2+r2)(p3+r3) }a_p^* a_r e^{ i(p +r).x}
[/tex]

But i'm sort of expecting the x integral to give some kind of delta function in r so that the r integral is easy leaving just the p integral ie the required result.
EDIT: maybe the commutation relations for the ap and bp operators will give delta functions, now to get the x integral out of the way first!
EDIT2: corrected sign error in exponential's.

TSny
TSny is offline
#4
Nov14-12, 02:14 PM
HW Helper
Thanks ∞
PF Gold
P: 4,513

Deriving charge for Noether current in free complex scalar field QFT


The delta function will come in through a well known "identity" that you can find here http://dlmf.nist.gov/1.17 (see equation 1.17.12).

Before going further, make sure you have the correct signs in your exponential functions for ##\pi## and ##\pi^*## (that you used to construct ##Q##).
Azelketh
Azelketh is offline
#5
Nov14-12, 03:53 PM
P: 36
ah, I think its [tex] \psi [/tex] and [tex] \psi^* [/tex] that have the incorrect signs.
Will edit through all above to correct this.

Bloody sign errors, when will they end!
Azelketh
Azelketh is offline
#6
Nov14-12, 07:58 PM
P: 36
ahh finally solved it,
using
[tex]
\int \partial^3 x e^{i(p+r).x} = (2\pi)^3 \delta^3(p+r) [/tex]
and working through.

Thanks for the assitance TSny!


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